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        <h3 id="一、题目"><a href="#一、题目" class="headerlink" title="一、题目"></a>一、题目</h3><p>    给定一个字符串，你的任务是计算这个字符串中有多少个回文子串。</p>
<p>    具有不同开始位置或结束位置的子串，即使是由相同的字符组成，也会被视作不同的子串。</p>
<span id="more"></span>

<p> <strong>示例 1：</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入：&quot;abc&quot;</span><br><span class="line">输出：3</span><br><span class="line">解释：三个回文子串: &quot;a&quot;, &quot;b&quot;, &quot;c&quot;</span><br></pre></td></tr></table></figure>

<p><strong>示例 2：</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入：&quot;aaa&quot;</span><br><span class="line">输出：6</span><br><span class="line">解释：6个回文子串: &quot;a&quot;, &quot;a&quot;, &quot;a&quot;, &quot;aa&quot;, &quot;aa&quot;, &quot;aaa&quot;</span><br></pre></td></tr></table></figure>

<p> <strong>提示：</strong></p>
<ul>
<li>  输入的字符串长度不会超过 1000 。</li>
</ul>
<h3 id="二、思路与代码"><a href="#二、思路与代码" class="headerlink" title="二、思路与代码"></a>二、思路与代码</h3><p>    首先我最先想到思路是<strong>递归</strong>，先计算开始位置的字符和它后面的字符能否构成回文子串及构成的个数，然后开始的位置递归后移，移动到末尾结束。</p>
<p>    时间复杂度比较大<code>O(n*n*n)</code>。</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br></pre></td><td class="code"><pre><span class="line">class Solution </span><br><span class="line">&#123;</span><br><span class="line">public:</span><br><span class="line">    int ans &#x3D; 0;</span><br><span class="line">    int countSubstrings(string s) </span><br><span class="line">    &#123; </span><br><span class="line">        if (s.empty())</span><br><span class="line">            return 0;</span><br><span class="line">        help(s, 0);  </span><br><span class="line">        return ans;</span><br><span class="line">    &#125;</span><br><span class="line">    void help(string&amp; s, int start)</span><br><span class="line">    &#123;</span><br><span class="line">        if (start &gt;&#x3D; s.size())</span><br><span class="line">        &#123;</span><br><span class="line">            return;</span><br><span class="line">        &#125;</span><br><span class="line">        int begin &#x3D; start;</span><br><span class="line">        for (int last &#x3D; begin; last &lt; s.size(); ++last)</span><br><span class="line">        &#123;</span><br><span class="line">            int i &#x3D; last;</span><br><span class="line">            while (begin &lt;&#x3D; i &amp;&amp; s[begin] &#x3D;&#x3D; s[i])</span><br><span class="line">            &#123;</span><br><span class="line">                if (begin &#x3D;&#x3D; i || begin + 1 &#x3D;&#x3D; i)</span><br><span class="line">                &#123;</span><br><span class="line">                    ++ans;</span><br><span class="line">                    break;</span><br><span class="line">                &#125;</span><br><span class="line">                --i;</span><br><span class="line">                ++begin;</span><br><span class="line">            &#125;</span><br><span class="line">            begin &#x3D; start;</span><br><span class="line">        &#125;</span><br><span class="line">        help(s, start + 1);</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<p>然后便是<strong>中心扩展法</strong>：</p>
<p>    这是一个比较巧妙的方法，实质的思路和动态规划的思路类似。</p>
<p>    比如对一个字符串 ababa，选择最中间的 a 作为中心点，往两边扩散，第一次扩散发现 left 指向的是 b，right 指向的也是 b，所以是回文串，继续扩散，同理 ababa 也是回文串。</p>
<p>    这个是确定了一个中心点后的寻找的路径，然后我们只要寻找到所有的中心点，问题就解决了。</p>
<p>    中心点一共有多少个呢？看起来像是和字符串长度相等，但你会发现，如果是这样，上面的例子永远也搜不到 abab，想象一下单个字符的哪个中心点扩展可以得到这个子串？似乎不可能。所以中心点不能只有单个字符构成，还要包括两个字符，比如上面这个子串 abab，就可以有中心点 ba 扩展一次得到，所以最终的中心点由 2 * len - 1 个，分别是 len 个单字符和 len - 1 个双字符。</p>
<p>    如果上面看不太懂的话，还可以看看下面几个问题：</p>
<ol>
<li><p>为什么有 2 * len - 1 个中心点？</p>
<ul>
<li><p>  aba 有5个中心点，分别是 a、b、c、ab、ba</p>
</li>
<li><p>  abba 有7个中心点，分别是 a、b、b、a、ab、bb、ba</p>
</li>
</ul>
</li>
<li><p>什么是中心点？</p>
<ul>
<li>  中心点即 left 指针和 right 指针初始化指向的地方，可能是一个也可能是两个</li>
</ul>
</li>
<li><p>为什么不可能是三个或者更多？</p>
<ul>
<li>  因为 3 个可以由 1 个扩展一次得到，4 个可以由两个扩展一次得到</li>
</ul>
</li>
</ol>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span></span><br><span class="line"><span class="class">&#123;</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="keyword">int</span> <span class="title">countSubstrings</span><span class="params">(string s)</span> </span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> n = s.<span class="built_in">size</span>(), ans = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; <span class="number">2</span> * n - <span class="number">1</span>; ++i) </span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">int</span> l = i / <span class="number">2</span>;</span><br><span class="line">            <span class="keyword">int</span> r = i / <span class="number">2</span> + i % <span class="number">2</span>;</span><br><span class="line">            <span class="keyword">while</span> (l &gt;= <span class="number">0</span> &amp;&amp; r &lt; n &amp;&amp; s[l] == s[r]) </span><br><span class="line">            &#123;</span><br><span class="line">                --l;</span><br><span class="line">                ++r;</span><br><span class="line">                ++ans;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> ans;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<p>还可以使用<strong>动态规划</strong>来进行解决：</p>
<ul>
<li>  状态：dp[i][j] 表示字符串s在[i,j]区间的子串是否是一个回文串。</li>
<li>  状态转移方程：当 s[i] == s[j] &amp;&amp; (j - i &lt; 2 || dp[i + 1][j - 1]) 时，dp[i][j]=true，否则为false</li>
</ul>
<p>这个状态转移方程是什么意思呢？</p>
<ul>
<li>  当只有一个字符时，比如 a 自然是一个回文串。</li>
<li>  当有两个字符时，如果是相等的，比如 aa，也是一个回文串。</li>
<li>  当有三个及以上字符时，比如 ababa 这个字符记作串 1，把两边的 a 去掉，也就是 bab 记作串 2，可以看出只要串2是一个回文串，那么左右各多了一个 a 的串 1 必定也是回文串。所以当 s[i]==s[j] 时，自然要看 dp[i+1][j-1] 是不是一个回文串。</li>
</ul>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span></span><br><span class="line"><span class="class">&#123;</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="keyword">int</span> <span class="title">countSubstrings</span><span class="params">(string s)</span> </span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> strLength = s.<span class="built_in">size</span>();</span><br><span class="line">        <span class="keyword">int</span> ans = <span class="number">0</span>;</span><br><span class="line">        vector&lt;vector&lt;<span class="keyword">bool</span>&gt;&gt; <span class="built_in">dp</span>(strLength, vector&lt;<span class="keyword">bool</span>&gt;(strLength));</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> j = <span class="number">0</span>; j &lt; s.<span class="built_in">length</span>(); j++) </span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt;= j; i++) </span><br><span class="line">            &#123;</span><br><span class="line">                <span class="keyword">if</span> (s[i] == s[j] &amp;&amp; (j - i &lt; <span class="number">2</span> || dp[i + <span class="number">1</span>][j - <span class="number">1</span>])) </span><br><span class="line">                &#123;</span><br><span class="line">                    dp[i][j] = <span class="literal">true</span>;</span><br><span class="line">                    ans++;</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> ans;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>



<p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/palindromic-substrings/solution/liang-dao-hui-wen-zi-chuan-de-jie-fa-xiang-jie-zho/">两道回文子串的解法（详解中心扩展法） - 回文子串 - 力扣（LeetCode） (leetcode-cn.com)</a></p>

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